91Ó°ÊÓ

Null hypotheses is,

${\mathrm{H}}_0:\mathrm{p}â‰\yen 0.13$

Alternative hypotheses is,

${\mathrm{H}}_0:\mathrm{p}<0.13$

The proportion of Americans who have seen or sensed angels is the random variable${\mathrm{P}}^{\text{'}}$.

Normal distribution with parameters $\mathrm{p}=0.13$and $\sqrt{\frac{\mathrm{p}(1-\mathrm{p})}{\mathrm{n}}}=\sqrt{\frac{0.13(1-0.13)}{76}}$.

The test statistic for this test is,

$\mathrm{z}=\frac{\left|{\mathrm{p}}^{\text{'}}-\mathrm{p}\right|}{\sqrt{\mathrm{p}(1-\mathrm{p})/\mathrm{n}}}$

$=\frac{0.026-0.13}{\sqrt{0.113/76}}$

When the p-value is bigger than the established alpha value, we do not reject the null hypothesis. Hence,

$\mathrm{p}\text{-value}=0.0036<0.05$

$=\mathrm{Î\pm }$

There the data to tell is that the percentage of Americans who have seen or sensed an angel is less than $13\%$at a level of significance of $5\%$.

${\mathrm{p}}^{\text{'}}-{\mathrm{z}}_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{{\mathrm{p}}^{\text{'}}\left(1-{\mathrm{p}}^{\text{'}}\right)}{\mathrm{n}}}≤\mathrm{p}≤{\mathrm{p}}^{\text{'}}+{\mathrm{z}}_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{{\mathrm{p}}^{\text{'}}\left(1-{\mathrm{p}}^{\text{'}}\right)}{\mathrm{n}}}........1$

percentage point $\frac{\mathrm{Î\pm }}{2}$of the standard normal distribution.

$\frac{\mathrm{Î\pm }}{2}=0.025$

$⇒{\mathrm{z}}_{\frac{\mathrm{Î\pm }}{2}}={\mathrm{z}}_{0.025}$

localid="1649942557746" $=1.96............2$

From equation $1$and$2$

$0.026-1.96\sqrt{\frac{0.026(1-0.026)}{76}}≤\mathrm{p}≤0.026+1.96\sqrt{\frac{0.026(1-0.026)}{76}}$

$0.026-1.96×0.018≤\mathrm{p}≤0.026+1.96×0.018$

Population proportion is,

$0≤\mathrm{p}≤0.623$

Test is,

$$\begin{gathered} \text{p-value}=\mathrm{P}\left\{{\mathrm{p}}^{\text{'}}<0.026\right\} \\ =\mathrm{P}\{\mathrm{z}<-2.688\} \\ =0.0036 \end{gathered}$$